Optimal. Leaf size=24 \[ \frac {e^x \left (-\frac {11}{2}+4 x\right ) (1+x-\log (4))}{5 x} \]
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Rubi [A] time = 0.10, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 9, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2199, 2176, 2194, 2177, 2178} \begin {gather*} \frac {4 e^x x}{5}-\frac {4 e^x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2177
Rule 2178
Rule 2194
Rule 2199
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \frac {e^x \left (11-11 x+5 x^2+8 x^3+\left (-11+11 x-8 x^2\right ) \log (4)\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (8 e^x x+5 e^x \left (1-\frac {8 \log (4)}{5}\right )-\frac {11 e^x (-1+\log (4))}{x^2}+\frac {11 e^x (-1+\log (4))}{x}\right ) \, dx\\ &=\frac {4}{5} \int e^x x \, dx+\frac {1}{10} (5-8 \log (4)) \int e^x \, dx+\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x^2} \, dx-\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x} \, dx\\ &=\frac {4 e^x x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x}-\frac {11}{10} \text {Ei}(x) (1-\log (4))-\frac {4 \int e^x \, dx}{5}+\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x} \, dx\\ &=-\frac {4 e^x}{5}+\frac {4 e^x x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 22, normalized size = 0.92 \begin {gather*} \frac {e^x (-11+8 x) (1+x-\log (4))}{10 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 26, normalized size = 1.08 \begin {gather*} \frac {{\left (8 \, x^{2} - 2 \, {\left (8 \, x - 11\right )} \log \relax (2) - 3 \, x - 11\right )} e^{x}}{10 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 35, normalized size = 1.46 \begin {gather*} \frac {8 \, x^{2} e^{x} - 16 \, x e^{x} \log \relax (2) - 3 \, x e^{x} + 22 \, e^{x} \log \relax (2) - 11 \, e^{x}}{10 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 27, normalized size = 1.12
method | result | size |
gosper | \(-\frac {{\mathrm e}^{x} \left (16 x \ln \relax (2)-8 x^{2}-22 \ln \relax (2)+3 x +11\right )}{10 x}\) | \(27\) |
risch | \(-\frac {{\mathrm e}^{x} \left (16 x \ln \relax (2)-8 x^{2}-22 \ln \relax (2)+3 x +11\right )}{10 x}\) | \(27\) |
norman | \(\frac {\left (\frac {11 \ln \relax (2)}{5}-\frac {11}{10}\right ) {\mathrm e}^{x}+\left (-\frac {8 \ln \relax (2)}{5}-\frac {3}{10}\right ) x \,{\mathrm e}^{x}+\frac {4 \,{\mathrm e}^{x} x^{2}}{5}}{x}\) | \(32\) |
default | \(-\frac {11 \,{\mathrm e}^{x}}{10 x}+\frac {4 \,{\mathrm e}^{x} x}{5}-\frac {3 \,{\mathrm e}^{x}}{10}-\frac {8 \,{\mathrm e}^{x} \ln \relax (2)}{5}+\frac {11 \ln \relax (2) {\mathrm e}^{x}}{5 x}\) | \(33\) |
meijerg | \(-\frac {11}{10 x}-\frac {3}{10}+\frac {11 \ln \relax (x )}{10}+\frac {11 i \pi }{10}+\frac {\frac {11 x}{10}+\frac {11}{10}}{x}-\frac {11 \,{\mathrm e}^{x}}{10 x}-\frac {11 \ln \left (-x \right )}{10}-\frac {11 \expIntegralEi \left (1, -x \right )}{10}-\left (-\frac {8 \ln \relax (2)}{5}+\frac {1}{2}\right ) \left (1-{\mathrm e}^{x}\right )-\left (-\frac {11 \ln \relax (2)}{5}+\frac {11}{10}\right ) \left (\ln \relax (x )+i \pi -\ln \left (-x \right )-\expIntegralEi \left (1, -x \right )\right )-\frac {2 \left (-2 x +2\right ) {\mathrm e}^{x}}{5}+\frac {11 \ln \relax (2) \left (\frac {1}{x}+1-\ln \relax (x )-i \pi -\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegralEi \left (1, -x \right )\right )}{5}\) | \(139\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.41, size = 44, normalized size = 1.83 \begin {gather*} \frac {4}{5} \, {\left (x - 1\right )} e^{x} + \frac {11}{5} \, {\rm Ei}\relax (x) \log \relax (2) - \frac {8}{5} \, e^{x} \log \relax (2) - \frac {11}{5} \, \Gamma \left (-1, -x\right ) \log \relax (2) - \frac {11}{10} \, {\rm Ei}\relax (x) + \frac {1}{2} \, e^{x} + \frac {11}{10} \, \Gamma \left (-1, -x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.07, size = 19, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (8\,x-11\right )\,\left (x-\ln \relax (4)+1\right )}{10\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 27, normalized size = 1.12 \begin {gather*} \frac {\left (8 x^{2} - 16 x \log {\relax (2 )} - 3 x - 11 + 22 \log {\relax (2 )}\right ) e^{x}}{10 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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